ExerciseODEsintermediate

Exercise: second-order ODE as a system, by hand

Reduce y''−sin y=0 to a first-order system and approximate y(3) with two Euler steps and one vector RK4 step.

Reduction to a system

Consider $y''-\sin y=0$ with $y(1)=2$ and $y'(1)=0$ , and approximate $y(3)$ . With the variables $y_1=y$ , $y_2=y'$ of the standard reduction, the system and its initial condition are:

Y'=F(t,Y)=\begin{bmatrix} y_2 \\ \sin y_1 \end{bmatrix},\qquad Y(1)=\begin{bmatrix} 2 \\ 0 \end{bmatrix}

Two vector Euler steps

ExampleEuler with h=1

Apply two explicit Euler steps with $h=1$ to the system, from $Y_0=[2,0]^T$ .

First step: $F(1,[2,0]^T)=[0,\ \sin 2]^T=[0,\ 0.9093]^T$ .
$Y_1=\begin{bmatrix}2\\0\end{bmatrix}+1\cdot\begin{bmatrix}0\\0.9093\end{bmatrix}=\begin{bmatrix}2\\0.9093\end{bmatrix}$
Second step: $F(2,[2,\ 0.9093]^T)=[0.9093,\ \sin 2]^T=[0.9093,\ 0.9093]^T$ .
$Y_2=\begin{bmatrix}2\\0.9093\end{bmatrix}+\begin{bmatrix}0.9093\\0.9093\end{bmatrix}=\begin{bmatrix}2.9093\\1.8186\end{bmatrix}$

The first component approximates the solution: $y(3)\approx 2.9093$ .

One vector RK4 step

ExampleRK4 with h=2

Apply one RK4 step with $h=2$ to the same system, from $Y_0=[2,0]^T$ at $t_0=1$ .

$K_1=F(1,[2,\ 0]^T)=[0,\ 0.9093]^T$ .
$K_2=F\bigl(2,\ [2,0]^T+1\cdot K_1\bigr)=F(2,[2,\ 0.9093]^T)=[0.9093,\ 0.9093]^T$ .
$K_3=F\bigl(2,\ [2,0]^T+1\cdot K_2\bigr)=F(2,[2.9093,\ 0.9093]^T)=[0.9093,\ 0.2302]^T$ , because $\sin 2.9093=0.2302$ .
$K_4=F\bigl(3,\ [2,0]^T+2\cdot K_3\bigr)=F(3,[3.8186,\ 0.4604]^T)=[0.4604,\ -0.6265]^T$ .
Final combination component by component:
$Y_1=Y_0+\frac{2}{6}\bigl(K_1+2K_2+2K_3+K_4\bigr)=\begin{bmatrix}3.3659\\0.8540\end{bmatrix}$

RK4 gives $y(3)\approx 3.3659$ (and $y'(3)\approx 0.8540$ ). Once more the high-order method substantially corrects Euler's estimate at the same evaluation cost.